Question

Every year on Groundhog Day (February 2), the famous groundhog Punxsutawney Phil tries to predict whether there will be 6 more weeks of winter. The article "Groundhog Has Been Off Target" (USA Today, Feb. 1,2011 ) states that "based on weather data, there is no predictive skill for the groundhog." Suppose that you plan to take a random sample of 20 years and use weather data to determine the proportion of these years the groundhog's prediction was correct. a. Describe the shape, center, and spread of the sampling distribution of \(\hat{p}\) for samples of size 20 if the groundhog has only a \(50-50\) chance of making a correct prediction. b. Based on your answer to Part (a), what sample proportion values would convince you that the groundhog's predictions have a better than \(50-50\) chance of being correct?

Short answer

The sampling distribution for a 50-50 chance is approximately normally distributed with a mean of 0.5 and standard deviation of 0.112. Sample proportions greater than 0.612 or less than 0.388 would indicate a predictive ability better than random chance (50-50 chance).

Step-by-step solution

Explore the Shape, Center, and Spread for a 50-50 chance prediction

In a binomial situation where the groundhog has a 50-50 chance, the distribution of the sample proportion \(\hat{p}\) would be close to the shape of a normal distribution. The Central Limit Theorem states that as the sample size gets larger, the shape of the distribution tends to approach a normal distribution shape. The center (mean of the distribution) would be the true proportion \(p=0.5\). The spread (standard deviation) of the distribution can be calculated using the formula \(\sqrt{ [ p(1-p) ] / n }\), where \(p\) is the proportion and \(n\) is the sample size. In our case this would be \(\sqrt{ [0.5(1-0.5)] / 20 } = 0.112\).

Evaluate Sample Proportions for better than 50-50 chance

To determine if the groundhog's predictions are better than random chance, we would need to observe sample proportions significantly higher than 0.5. The boundary for significance can be evaluated using z-scores. Any z-score greater than 1 or less than -1 is usually considered significant. Using the formula \((\hat{p} - p) / \sqrt{ [ p(1-p) / n ] }\), where \(\hat{p}\) is the sample proportion, we get \( (\hat{p} - 0.5) / 0.112 \). So, any \(\hat{p}\) that gives a Z-Score > 1 or < -1 would be considered significant. Solving this gives \(\hat{p} > 0.612\) or \(\hat{p} < 0.388\) are the values which would indicate predictions better than a 50-50 chance.

Key concepts

Binomial Distribution

The binomial distribution is a fundamental probability distribution that describes the number of successes in a fixed number of independent trials, with the same probability of success on each trial. It applies perfectly to the Groundhog Day exercise, where each year's prediction can be seen as a separate trial with two possible outcomes: correct (success) or incorrect (failure). The key parameters of a binomial distribution are

• The number of trials (n), which in our case is 20 years.
• The probability of success on any given trial (p), which is assumed to be 0.5 given the groundhog has a 50-50 chance of making a correct prediction.

When we talk about the shape of a binomial distribution, it can be symmetric or skewed depending on the value of p. If p is equal to 0.5, the distribution is symmetric, resembling a normal distribution when n is large. This symmetry is what we are utilizing in our Groundhog Day example.

Central Limit Theorem

The Central Limit Theorem (CLT) is a powerful statistical concept that explains why the binomial distribution of the sample proportion \(\hat{p}\) approximates a normal distribution as the sample size increases. It states that regardless of the distribution of the population, the sampling distribution of the mean will tend to be normal if the sample size is sufficiently large. In simpler terms, if you repeat a sampling process multiple times, the means of these samples will form a distribution that is approximately normal.

This theory is pivotal in predicting the groundhog's performance if we consider each year as a sample. Assuming we have enough samples (in our case, 20 years is considered adequate), we can expect the sample proportions to follow a normal distribution, making it easier to apply further statistical analysis, such as calculating z-scores to test predictions.

Sample Proportion

In statistics, the sample proportion is a ratio that gives us the fraction of times an event occurs in our sample. For the Groundhog Day problem, \(\hat{p}\) represents the proportion of years in which the groundhog's predictions were accurate among the sampled years.

To assess the sampling distribution of \(\hat{p}\), we consider it as a mean of a binomial distribution for our 20 selected years. The calculation of the sample proportion involves counting the number of successes (i.e., correct predictions) and dividing by the total number of trials (i.e., total years sampled). This proportion helps us determine whether or not there is evidence to suggest that the groundhog has predictive skill, based on the observed frequency of correct predictions within the sample.

Z-Score

A z-score, also known as a standard score, provides a way of describing the position of a raw score in terms of its distance from the mean, measured in standard deviations. It's a measure that helps in understanding how extraordinary a value is within a set of data. To calculate a z-score, you subtract the mean from the raw score and then divide by the standard deviation.

In our context, once we have the sample proportion from the Groundhog Day predictions, we can use its z-score to determine how likely it is due to random chance. A z-score above 1 or below -1 generally suggests that the observed proportion is significantly different from the hypothesized probability of 0.5, indicating that the groundhog's predictions are not merely due to chance. For instance, if the proportion of correct predictions leads to a z-score greater than 1, we could say that the groundhog has predictive skill better than flipping a coin.